Finding the probability that each child gets at least 1 ball when we are distributing 5 DISTINCT balls among 4 children(who are distinct of course). My approach:-
First, I found the total number of cases(the sample space). We can find that out from, $$4^5=1024$$
Now for favourable cases, I did this:-
I selected any 4 balls out of the given 5 by $${5\choose 4}$$ then I gave each child 1 ball out of the selected 4 balls, that can be done in 4!  ways. With this we have provided each child 1 ball hence satisfying the condition given the question that no one should go home empty handed. Now we have 1 ball left which has to be given to any of the 4 children. So that ball can be given in 5 ways.
Hence the total favourable cases become, $${5\choose4}(4!)(5)=600$$ So the probability becomes $$600/1024$$But the answer given in the book is 15/64.
 A: You’re overcounting, as described in aryan bansal’s comment. A valid method would be to use Inclusion-Exclusion for counting the number of ways atleast one child gets no ball, which is $$S={4\choose 1} 3^5 -{4\choose 2} 2^5 +{4\choose 3} 1^5 $$ and then subtract this from the total ways, giving the number of ways where every child gets atleast one ball. The answer therefore is $$ \frac{4^5-S}{4^5} =\frac{15}{64}$$
A: Since there are four possible recipients for each of the five balls, your denominator is correct.
As for the numerator, notice that if each child receives at least one ball, then one child must receive two of the five balls, while each of the others receives one ball.  There are four ways to select the child who will receive two balls, $\binom{5}{2}$ ways to select which two of the five balls that child will receive, and $3!$ ways to distribute the remaining three distinct balls to the other three children so that each of those children receives one ball.  Hence, the number of favorable cases is
$$\binom{4}{1}\binom{5}{2}3!$$
Therefore, the probability that each child receives at least one ball when five distinct balls are randomly distributed to four children is
$$\frac{\binom{4}{1}\binom{5}{2}3!}{4^5} = \frac{240}{1024} = \frac{15}{64}$$
What errors did you make?

Now we have $1$ ball left which has to be given to any of the $4$ children. So that ball can be given in $\color{red}{5}$ ways.

There are four possible recipients for that ball, so it can only be distributed in four ways.  Had you not made that mistake, your count would have been
$$\binom{5}{4}4!4 = 480$$
which is exactly twice the number of favorable cases.
Why?
As aryan bansal pointed out in the comments, your method counts each favorable distribution twice, once for each way of designating one of the two balls the child who receives two balls as the ball that child receives and the other as the additional ball.  For instance, you count the following distribution twice:  Anna receives a pink ball and a yellow ball, Brenda receives a red ball, Charles receives a blue ball, and Dana receives a green ball.  You count it once when you designate the pink ball as the ball Anna must receive and the yellow ball as the additional ball, and once when you designate the yellow ball as the ball Anna must receive and the pink ball as the additional ball.
A: Would a stars and bars approach be appropiate for this problem? I mean, we're placing n=5 objects (balls) into k=4 bins (children)... Just an idea
